Question

\[\text{If }\dfrac{\tanh x+1}{\tanh x-1}=-{{e}^{px}},\text{ then }p=\]
(a) 1
(b) 2
(c) 3
(d) 0

Answer 1

Hint: By using the known formula of sinh x, cosh x, find out the value of tan hx. Then convert the whole equation into exponential terms. Now, take the least common multiple and cancel the common terms. Then, use the properties of powers to simplify the equation. Compare it to the given value to find the value of p which is our required result.

Complete step-by-step answer:
The given expression in the question can be written in the form:
\[\dfrac{\tanh x+1}{\tanh x-1}\]
By basic knowledge of the hyperbolic functions, we know that
\[\tanh x=\dfrac{\sinh x}{\cosh x}\]
By substituting this into given expression, we get it as:
\[\dfrac{\dfrac{\sinh x}{\cosh x}+1}{\dfrac{\sinh x}{\cosh x}-1}\]
By taking the least common multiple in the numerator, we get it as:
\[\dfrac{\dfrac{\sinh x+\cosh x}{\cosh x}}{\dfrac{\sinh x}{\cosh x}-1}\]
By taking the least common and canceling common terms, we get,
\[\dfrac{\sinh x+\cosh x}{\sinh x-\cosh x}\]
From the above terms, we get the equation as in the form of:
\[\dfrac{\tanh x+1}{\tanh x-1}=\dfrac{\sinh x+\cosh x}{\sinh x-\cosh x}\]
By basic knowledge of hyperbolic function, we know the formula,
\[\sinh x=\dfrac{{{e}^{x}}-{{e}^{-x}}}{2};\cosh x=\dfrac{{{e}^{x}}+{{e}^{-x}}}{2}\]
By substituting these into our expression, we get it as:
\[\dfrac{\dfrac{{{e}^{x}}-{{e}^{-x}}}{2}+\dfrac{{{e}^{x}}+{{e}^{-x}}}{2}}{\dfrac{{{e}^{x}}-{{e}^{-x}}}{2}-\dfrac{{{e}^{x}}+{{e}^{-x}}}{2}}\]
By taking the least common multiple and canceling 2, we get it as,
\[\dfrac{{{e}^{x}}-{{e}^{-x}}+{{e}^{x}}+{{e}^{-x}}}{{{e}^{x}}-{{e}^{-x}}-{{e}^{x}}-{{e}^{-x}}}\]
By canceling the common terms, we get the expression as
\[\dfrac{2{{e}^{x}}}{-2{{e}^{-x}}}\]
By canceling 2, we get the expression in the form of:
\[-\dfrac{{{e}^{x}}}{{{e}^{-x}}}\]
By basic knowledge of power, we can say the relation:
\[\dfrac{{{a}^{b}}}{{{a}^{c}}}={{a}^{b-c}}\]
By applying that here, we get the expression in the form of:
\[-{{e}^{x-\left( -x \right)}}\]
By simplifying it, we can write it as follows:
\[-{{e}^{2x}}\]
By comparing this to \[-{{e}^{px}}\], we can say that the value of p to be:
p = 2
Therefore, option (b) is correct.

Note: Be careful while substituting sinh, cosh as the terms increases, there will be more risk of missing signs like “+” or “–“. So, point out and write every term with utmost care. While converting sin, cos into exponential you must end up with \[{{e}^{px}}\] in numerator, or else the value of p will become negative. So, do it carefully.